Generate samples from a normal distribution

Hello,

I have a Gaussian mixture distribution which looks as attached. I now want to sample and generate 1000 samples from this and plot their histogram. I tried using randi function but it does not work. Please let me know how to do this.

y = randi(my_pdf,100);  %%my_pdf is my distribution

Thanks!

[Walter Roberson]

Is that a continuous function or a discrete function?

[Kash022]

Continuous generated using normpdf

[Torsten]

Could you explain the graphic in gmm.jpg ?

It's not a distribution generated with normpdf, not even a distribution ...

So what is it ?

Best wishes

Torsten.

[Kash022]

Its a gaussian mixture distribution with different means. This is the code snippet used to generate it.

clear all;
XX= [-2:0.01:20];
some_rand_noise = 0.1; 
my_pdf = zeros(1,length(XX));
for k = 1:256
for ii=1:256

      val(ii) = hw(bitxor(ii-1,k-1))+hw(k-1); %%hw is a function used to generate hamming weights i.e no of ones %%%
      my_pdf = my_pdf + normpdf(XX,val(ii),some_rand_noise);
  end
  end 
      figure();plot(XX,my_pdf/256); hold on;

[Image Analyst]

Please supply the hw function so we can run your code.

Answer 1

[Image Analyst]

If you have that function already, then use inverse transform sampling: https://en.wikipedia.org/wiki/Inverse_transform_sampling. In short, compute the CDF of your function and use rand() to pick a value.

I attach an example of how I used it for the Rayleigh distribution.

Answer 2

[CarrotCakeIsYum]

randi will generate rando integers.

To generate a random sample from a vector, use randsample. see:

help randsample

I'd imagine you'd want:

y = randsample(my_pdf,100);

To plot a histogram use the 'hist' command.

[Kash022]

thanks! but how do I do this for different means? as you can see in my attachment, I have different means,for each of them I need to generate the samples and then plot their histogram. Thanks!

[CarrotCakeIsYum]

Sorry I don't follow... your population will only have one mean!

For the JPEG you've posted, the mean will be at 7.5 (since it is a symmetrical distribution).

Answer 3

[Image Analyst]

Maybe use randn() to get a list of a few million numbers taken from those 13 distributions (call randn thirteen times), then pick one of the values at random.